CONTINUITY |
For a function to be continuous, the rule of thumb is that you can draw the graph of this continuous function with a continuous line. That is, you don't have to pick up your pen as you draw the curve.
More formally, a function is discontinuous at vertical asymptotes and other values of x that don't belong to the domain of the function. Points where no y-value can be defined.
For example if , we know that f (x) is continuous on its domain of . This function is discontinuous when x > 5 since, in the set of Reals, we cannot find an even root for a negative number. So, f (x) is discontinuous on intervals where (x, f (x)) are not defined by the rule of correspondence.
Generally, when asked to find where a function is continuous, we find where it is discontinuous and say that it's continuous everywhere else.
There are 3 parts to the test for continuity.
To test if a function f(x) is continuous at the point x = a, test if
1) f (a) is defined.
2) exists.
3) .
What does each item in the list mean?
1) f (a) is defined means that we get a y-value when we substitute x = a.
2) for to exist, must equal .
3) f (a) = means the y-value of part 1 must equal the limit of part 2.
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Test the function for continuity at x = 1.
test part 1
f (1) = 2(1) – 1 = 1, so f (1) is defined.
test part 2
, but .
Since 1 doesn't = 2, f (x) is discontinuous at x = 1.
f (x) fails part 2 of the test for continuity.
Had the function been ,
it would be continuous at x = 1 since
both the left and right side limits would = 1 which equals f (1).
This function would pass all 3 parts of the continuity test.
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Example: On the curve below, test the continuity at x = 3:
f (3) = 2 so part 1 works.
, so part 2 works.
But 2 doesn't = 6, so part 3 doesn't work.
The y-value at "a" must equal the limit value at "a".
If we set f (3) = 6, then this function is continuous at x = 3.
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Classic Exam Question:
Given , find a value for c that makes f (x) continuous at x = 2.
First, we'll find f (2), then we'll find the limit as x approaches 2, and then we'll set them equal to satisfy the statements of the test for continuity.
f (2) = 4c – 3
now solve left limit = right limit:
4c – 3 = 2c + 2 gives us c = 5/2 or 2.5.
This makes f (2) = 7 and limit as x approaches 2 = 7.
In similar questions with 2 unknowns -- say c and d, we get enough information to find a system of 2 equations in order to solve for the 2 unknowns. In such cases, there will also be 2 different values to limit towards. See practice question 2b.
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1) Discuss the continuity of these functions:
a) | b) | c) |
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2) Determine which values of c and/or d make f (x) continuous on all Real numbers.
a) Given
b) Given
continuity intro | test for continuity | examples |
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1) Discuss the continuity of these functions:
a) f (x) continuous on |
b) g(x) continuous on (vertical asymptotes) | |
c) , here we have 3 situations to discuss. 1/ The domain of 2/ The domain of 3/ The restriction on (x – 4) in the denominator is h (x) continuous on [ – 5, – 3] [3, 4) (4, 5] interval notation: |
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2) Determine which values of c and/or d make f (x) continuous on all Real numbers.
a) Given
First, we'll find f(1), then we'll find the limit as x approaches 1, and then we'll set them equal to satisfy the statements of the test for continuity.
f (1) = 3c – 2
now solve left limit = right limit:
c^{ 2} = 3c – 2 becomes c^{ 2} – 3c + 2 = 0.
(c – 2)(c – 1) = 0
Thus, c can be either 2 or 1
If c = 2, f (1) = 4 and limit as x approaches 1 = 4.
If c = 1, f (1) = 1 and limit as x approaches 1 = 1.
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b) Given
Here there are 2 points of contention. The function changes rule at x = – 1 and x = 2.
We'll set up 2 equations in 2 unknowns (namely c and d), then solve for the common solution.
at x = – 1:
f ( – 1) = – 4 |
so ( – c + d = – 4 ) (eq_{1}: 1st equation in c and d)
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at x = 2:
f (2) = – 10 |
so 2c + d = – 10 (eq_{2}: 2nd equation in c and d)
Now, we solve the system:
Subtract eq_{1} from eq_{2} to get 3c = – 6, so c = – 2, then substitute for d = – 6.
Now,
f ( – 1) = – 4 | and |
f (2) = – 10 | and |
So f (x) is continuous on all Reals when c = – 2 and d = – 6.
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